A man reaches his office 30 min late, if he walks from his home at 3 km per hour and reaches 40 min early if he walks 4 km per hour. How far is his office from his house?
Lets assume distance between office and his house =x km
Time taken when speed is 3 kmph = x/3 min.
Time taken when speed is 4 kmph = x/4 min.
Difference between time taken 40-(-30) = 70mins = 70/60 hours.
x/3- x/4 = 70/60
x/12= 7/6
x=14 km
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City bus runs two bus from A to B each making 5 round trips in a day. There are no stops between. These buses ply back and forth on same route at diff. but uniform speed. Each bus start at 7 am from respective terminus. They meet for the first time at 7 km from A. Their next meeting is at distance of 4 km from B, while travelling in opposite direction. Assuming that time by buses at terminus is negligibly small, and cost of running a bus is 20 per km. Find daily cost of buses ?
Let's assume total distance travelled by both bus = x km
So as per question
First meet at 7 km from A.
By that time first bus travels 7 Km and second one travels x-7 Km. As they have travelled for the same time,
Second meeting is at 4 km from B:
By that time first bus travels (x+4) Km and second bus will travels x+(x-4) = (2x-4) km.
=> 7/(x+4) = (x-7)/(2x-4)
x^2 - 15x = 0 => x=17 km.
So one round trip = 2*17 = 34 km.
as given each bus make 5 round trips in a day.
so total distance cover by both buses = 34 * 2 *5 = 340 km.
Cost of travelling per bus = 340 * 20 = Rs. 6800.
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A man sets out to cycle from Delhi to Rohtak and at the same time another man starts from Rohtak to cycle to cycle to Delhi. After passing each other they completed their journey in (10/3) hours and (16/3) hours respectively.At what rate does the second man cycle if the first cycle at 8 kmph?
3/4 of a man's usual speed means, he takes 4/3 of his usual time to cover the same distance,
i.e. he takes 4/3 â€“ 1 = 1/3 time extra.
Given 1/3 time is 20 minutes
Usual time = 20 * 3 = 60 minutes.
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If a man walks at the rate of 5kmph, he misses a train by only 7min. However if he walks at the rate of 6 kmph he reaches the station 5 minutes before the arrival of the train.Find the distance covered by him to reach the station.
Lets assume the required distance = x km.
Difference in the times taken at two speeds=12mins=1/5 hr.
Therefore (x/5-x/6)=1/5 or (6x-5x) = 6 or x = 6km.
So required distance = 6 km
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Yana and Gupta leave points x and y towards y and x respectively simultaneously and travel in the same route. After meeting each other on the way, Yana takes 4 hours to reach her destination, while Gupta takes 9 hours to reach his destination. If the speed of Yana is 48 km/hr, what is the speed of Gupta?
Let the speed of Yana be v1 and Gupta be v2 and time taken by them be t1 and t2 respectively.
There is a shortcut for such questions as v1^2/v2^2= t2/t1
Putting the respective values we get v2= 32km/hr
1.6 km =1 miles
So 32/1.6= 20mph.
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Walking 5/6 of its usual speed, a train is 10min late. Find the usual time to cover the journey?
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